Absentee rates - Friday vs Wednesday: We want to test whether or not more studen
ID: 3217542 • Letter: A
Question
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 292 students with Friday afternoon classes, 58 missed the class. In a different random sample of 297 students with Wednesday afternoon classes, 34 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software.
Data Summary
SE = 0.02992
The Test: Test the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes. Use a 0.05 significance level.
(a) Letting p1 be the absentee rate from the sample on Friday and p2 be the rate from Wednesday, calculate the test statistic using software or the formulaz =
where p is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) given with the data. Round your answer to 2 decimal places.
z =
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.
(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places.
P-value =
(c) What is the conclusion regarding the null hypothesis?
reject H0
fail to reject H0
(d) Choose the appropriate concluding statement.
The data supports the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.
There is not enough data to support the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.
We have proven that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.
We reject the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.
total number total number Proportion Day of absences (x) of students (n) p = (x/n) Friday 58 292 0.19863 Wednesday 34 297 0.11448Explanation / Answer
Given that,
sample one, x1 =58, n1 =292, p1= x1/n1=0.199
sample two, x2 =34, n2 =297, p2= x2/n2=0.114
null, Ho: p1 = p2
alternate, more students are absent on Friday afternoon classes H1: p1 > p2
level of significance, = 0.05
from standard normal table,right tailed z /2 =
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.199-0.114)/sqrt((0.156*0.844(1/292+1/297))
zo =2.813
| zo | =2.813
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =2.813 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2.8127 ) = 0.00246
hence value of p0.05 > 0.00246,here we reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 2.813
critical value: 1.645
decision: reject Ho
p-value: 0.00246
we have evidence to support the claim, that more students are absent on Friday afternoon classes